poj

Cow Bowling

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 18716 Accepted: 12463

Description

The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this: 

          7        3   8      8   1   0    2   7   4   4  4   5   2   6   5

Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow's score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame. 

Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.

Input

Line 1: A single integer, N 

Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.

Output

Line 1: The largest sum achievable using the traversal rules

Sample Input

573 88 1 02 7 4 44 5 2 6 5

Sample Output

30

Hint

Explanation of the sample: 

          7         *        3   8       *      8   1   0       *    2   7   4   4       *  4   5   2   6   5

The highest score is achievable by traversing the cows as shown above.

这个题吧，基本紫书上有介绍，我就直接写出来了，居然一遍就过了

就是数字三角形那部分

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<algorithm>using namespace std;int dp[355][355];int a[355][355];int main(){int n,i,j;while(~scanf("%d",&n)){for(i=1;i<=n;i++){for(j=1;j<=i;j++){scanf("%d",&a[i][j]);}}for(i=1;i<=n;i++){dp[n][i]=a[n][i];}for(i=n-1;i>=1;i--){for(j=1;j<=i;j++){dp[i][j]=max(dp[i+1][j],dp[i+1][j+1])+a[i][j];}}printf("%d\n",dp[1][1]);memset(dp,0,sizeof(dp));}    return 0;}